Answer:
Heat energy required (Q) = 10.736 KJ
Explanation:
Given:
Specific heat of ethanol (C) = 2.44 J/g °C
Mass of ethanol (M) = 50 gram
Initial temperature (T1) = -20°C
Final temperature (T1) = 68°C
Find:
Heat energy required (Q) = ?
Computation:
Change in temperature (ΔT) = 68°C - (-20°C)
Change in temperature (ΔT) = 88°C
Heat energy required (Q) = mC(ΔT)
Heat energy required (Q) = (50)(2.44)(88)
Heat energy required (Q) = 10,736 J
Heat energy required (Q) = 10.736 KJ
Answer:
Losing 2 valence electrons
Gaining 2 electrons
SrSo4 = Sr(2+) + SO4(2-)
Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).
When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4
Answer:
The volume is increased.
Explanation:
According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;
V₁ / T₁ = V₂ / T₂ -----(1)
In statement the data given is,
T₁ = 10 °C = 283.15 K ∴ K = 273.15 + °C
T₂ = 20 °C = 293.15 K
So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,
V₁ = 100 L
V₂ = Unknown
Now, we will arrange equation 1 for V₂ as,
V₂ = V₁ × T₂ / T₁
Putting values,
V₂ = 100 L × 293.15 K / 283.15 K
V₂ = 103.52 L
Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.
Answer:
398 mL
Explanation:
Using the equation for molarity,
C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L
V₂ = V₁ + V' where V' = volume of water added.
So, From C₁V₁ = C₂V₂
V₁ = C₂V₂/C₁
= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L
= 0.875 mol/8.61 mol/L
= 0.102 L
So, V₂ = V₁ + V'
0.5 L = 0.102 L + V'
V' = 0.5 L - 0.102 L
= 0.398 L
= 398 mL
So, we need to add 398 mL of water to the nitric solution.
